LeetCode | Partition List

https://leetcode.com/problems/partition-list/

创建两个dummy node：leftDummy和rightDummy，分别对应划分后的两个链表。然后遍历原始链表，比给定的值小就连到leftDummy链表后面，否则就连到rightDummy链表后面。最后将两个链表连起来，注意将尾节点的next域置空。

Time complexity: O(n)

Space complexity: O(1)

C++

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        if (!head || !head->next) return head;
        
        ListNode left_dummy(0), right_dummy(0);
        ListNode *l = &left_dummy, *r = &right_dummy;
        while (head) {
            if (head->val < x) {
                l->next = head;
                l = l->next;
            } else {
                r->next = head;
                r = r->next;
            }
            head = head->next;
        }
        l->next = right_dummy.next;
        r->next = NULL;
        return left_dummy.next;
    }
};